Question
A function $$f$$ is defined as follows $$f\left( x \right) = {x^p}\cos \left( {\frac{1}{x}} \right),\,x \ne 0\,f\left( 0 \right) = 0$$
What conditions should be imposed on $$p$$ so that $$f$$ may be continuous at $$x = 0\,? $$
A.
$$p = 0$$
B.
$$p > 0$$
C.
$$p < 0$$
D.
No value of $$p$$
Answer :
$$p > 0$$
Solution :
Given function is defined as : \[f\left( x \right) = \left\{ \begin{array}{l}
{x^p}\cos \left( {\frac{1}{x}} \right),\,\,\,x \ne 0\\
\,\,\,\,\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0
\end{array} \right.\]
For continuity :
$$\eqalign{
& {\text{L}}{\text{.H}}{\text{.S}}{\text{.}}:\mathop {\lim }\limits_{x \to 0} f\left( x \right) = {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) \cr
& \Rightarrow \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} {x^p}\cos \left( {\frac{1}{x}} \right) = 0 \cr
& \Rightarrow \mathop {\lim }\limits_{x \to 0} {x^p}\cos \left( {\frac{1}{x}} \right) = 0 \cr} $$
$$\cos \left( {\frac{1}{x}} \right)$$ is always a finite quantity if $$x \to 0$$
$$ \Rightarrow {x^p} = 0$$
which is possible only if $$p > 0.$$